Let the three consecutive terms be$\frac{a}{r},a,ar$

Product =$\left(\frac{a}{r}\right)$(a)(ar) = $a^3$ = 512, $\Rightarrow$a = 8

Sum of these numbers$=a(1+r+\frac{1}{r})=42\Rightarrow 8(1+r+\frac{1}{r})=42$

$4r^2-17r+4=0$

(4r – 1)(r – 4) = 0

r = $\frac{1}{4}$or r = 4.

Hence the largest number is 32. The series is 32, 8, 2 or 2, 8, 32. 

The first two terms = a, ar

$a+ar=\frac{5}{3}$

$\frac{a}{(1-r)}=3$

$a=3(1-r)$

$a(1+r)=\frac{5}{3}$

$3(1-r)(1+r)=\frac{5}{3}$

$1-r^2=\frac{5}{9}$

$r^2=\frac{4}{9}$

$r=\frac{2}{3}$

$a=3(1-\frac{2}{3})=1$

First term is 1. 

Let the numbers be x and y.

$x+y=30$

xy = 144

$(x-y{)}^2=(x+y{)}^2-4xy$

$900-(4\times 144)=324$

 hence $(x–y)=18$

Solving:$x-y=18\&x+y=30$

x = 24,       y = 6

So, the bigger number is 24. 

$G.M=(A.M\times H.M{)}^{\frac{1}{2}}$
$64=10\times H.M$
$H.M=6.4$
 

$6x^2-4x+3x-2=0$
$2x(3x-2)+1(3x-2)=0$
$(2x+1)(3x-2)=0$
The roots are:$\frac{-1}{2},\frac{2}{3}$

Soln:

The$3^{rd}$ term will be$x+2d$

The${15}^{th}$ term will be$x+14d$

From the information given in the question,

$x+2d+x+14d=x+5d+x+10d+x+12d$

Solving we get,$x+11d=0$, this is nothing but the${12}^{th}$term of the series.

Let $x^3=y$

$y^2-26y-27=0$

$y^2-27y+y-27=0$

$y(y-27)+1(y-27)=0$

$(y+1)(y-27)=0$

The roots are y = -1, y = 27

$x^3=-1,x=-1$

$x^3=27,x=3$

$\left[\right(x+1\left)\right(x+4\left)\right]\left[\right(x+2\left)\right(x+3\left)\right]=24$

$(x^2+5x+4)(x^2+5x+6)=24$

Let$x^2+5x=k$

$(k+4)(k+6)=24$

$k^2+10k=0$

$k(k+10)=0$

$k=0,k=-10$

$x^2+5x=0,x^2+5x=-10$

$x(x+5)=0,x^2+5x+10=0$

$x=0,x=-5;D=25-40=-15$ No real roots.

So, roots are 0, -5.

Let $x-\frac{1}{x}=z$
$(x-\frac{1}{x}{)}^2=z^2$
$(x^2+\frac{1}{x^2}-2)=z^2$
$x^2+\frac{1}{x^2}=z^2+2$
Now substituting in the equation:
$6(z^2+2)-25z+12=0$
$6z^2+12-25z+12=0$
$6z^2-25z+24=0$
$6z^2-16z-9z+24=0$
$2z(3z-8)-3(3z-8)=0$
$(2z-3)(3z-8)=0$
$z=\frac{3}{2};z=\frac{8}{3}$
$x-\frac{1}{x}=\frac{3}{2}$
$x-\frac{1}{x}=\frac{8}{3}$
$2x^2=3x-2=0$
$3x^2-8x-3=0$
$2x^2-4x+x-2=0$
$3x^2-9x+x-3=0$
$2x(x-2)+1(x-2)=0$
$3x(x-3)+1(x-3)=0$
$(2x+1)(x-2)=0$
$(3x+1)(x-3)=0$
$x=-\frac{1}{2},x=2$
$x=-\frac{1}{3},x=3$
So,the roots are:$-\frac{1}{2},2,-\frac{1}{3},3$
 

$2^x{.2}^3+2^{-x}=6$
Let$2^x=z$
$8z+\frac{1}{z}=6$
$8z^2+1=6z$
$8z^2-6z+1=0$
$8z^2-4z-2z+1=0$
$4z(2z-1)-1(2z-1)=0$
$(4z-1)(2z-1)=0$
$z=\frac{1}{4},$
$z=\frac{1}{2}$
$2^x=\frac{1}{4}$
$x=-2$
$2^x=\frac{1}{2}$
$x=-1$
Roots are $x=-2,x=-1$